Post by: Eternal Student on 20/01/2022 17:30:31
What's this about?
I've been asked to help someone else with a physics question. It's a bit difficult and I think the wording in the question isn't great. I would not like to advise them badly, so I would appreciate it if some other person(s) who feel they would know the topic would also have a look at it.
What are the complications and consequences?
The question comes from a past paper that was set by a UK university for their 1st year Physics undergradutes. Technically a small portion of the marks would end up contributing to the students final award etc.
There might be copyright issues, for example the university may not be happy about their material being printed here. It's quite common for universities to recycle a few questions and use them again some years later etc.
It might not be in the best interests of their 1st year undergrads if we just have a complete answer and discussion printed here. If the text is exactly the same as the original question then Google will find it and some students might accidentally deprive themselves of the opportunity to learn by working on the problem themselves.
Mitigating circumstances: This is from a past paper (last year's exam to be specific). Hence, the question is already "out there" and the marking scheme (almost a model answer) has already been circulated by the University to this year's students who were using the past paper purely for revision or practice.
What was I thinking of doing?
Re-print parts of the question here. I may alter one or two small details but I don't want to alter too much of the wording because I think it's the wording that is half the problem.
What am I seeking now?
Guidance from a moderator or similar. Is it better to just steer clear of re-printing anything that might cause harm or breech some copyright of another university? Especially since I'm inclined to criticise the wording of the question.
Best Wishes.
Post by: Halc on 20/01/2022 18:08:13
The person needing help needs to show effort: An attempt to solve the problem is posted, and then the site can help point out where the things are going wrong. But at no point should an answer just be posted.
It gets more complicated if you're requesting help to a person who is not on the site, but your help to them should still proceed along the above guidelines.
About copyright issues, I've never seen issues with it. I've seen scans of the actual problem in the book.
Post by: alancalverd on 20/01/2022 21:02:45
Hence, the question is already "out there" and the marking scheme (almost a model answer) has already been circulated by the University to this year's students who were using the past paper purely for revision or practice.As it's in the public domain I don't see a problem in principle - you aren't intending to profit from reproducing the contentious part, and there's a public interest case for discussing it openly.
Post by: Eternal Student on 20/01/2022 22:50:33
Late editing: I don't have a better copy to read from myself - I would just be typing out what I think should be there (for example the superscripts in the usual relativistic mass formula).
Overview of the exam question:
The marks are shown in [bracketts] at the end of each section.
To complete the exam in time you should be trying to average a mark a minute and certainly less than 2 minutes per mark.
My concern is the last part of the question, section (g).
In my view the wording of the question is extremely poor, it could take 5 minutes just to make sense of it.
If anyone has 10 minutes spare, would you like to try section (g) and see what you think? It does refer to earlier parts of the question, so in all honesty you would probably want to skim through the whole question.
Here's the question for section (g) typed out with as much of the original punctuation as I can preserve:
(g) If you could use the same experiment described in the first part of the question to measure the mass of protons moving at 0.2c, how many experiments would you need to average in order to detect the relativistic mass change at the 95% confidence level in the mean? [5 marks]
Here's the main description of the experiment from the start of the question:
1. An experiment to measure the mass of a proton at rest yields measurements that are distributed as a Gaussian with standard deviation σ = 0.2 x 10 -27 Kg.
Minor note: I've probably already helped too much by filtering out most of the incidental irrelevant junk in the question but let's just continue to do this anyway:
If you scan through the whole of the question, you will also have this useful information:
Earlier questions involved relativistic mass. You have values given for the rest mass of a proton (M0 = 1.7 x 10 -27 Kg ) and you will have calculated that the mass of a proton at 0.2c is 2% larger than the rest mass.
- - - - -
That's it. Try section (g) now.
Post by: Eternal Student on 21/01/2022 06:11:28
So this is the first critical comment I would have about the question:
It does not matter how fast a proton is travelling, its mass is an invariant. If a student answers part (g) by assuming that the mass does NOT change regardless of the speed of the proton then it is technically correct. Just to be clear, the precise wording in section (g) stated that the experiment could be used to measure "mass" it did not state that the experiment could be used to determine "relativistic mass". How you could use this to detect changes in relativistic mass is your own problem to think about.
Students can lose some minutes here. In these modern times, they should not be faulted for deciding that "mass" means "rest mass". Eventually you might decide that the experiment can be applied to determine relativistic mass directly AND that the distribution and standard deviation of the measurements will be unchanged.
Another reasonable course of action the student could take is to use the relativistic mass formula given earlier in the question and/or the simplified first order series they were asked to find. If you read the question carefully you should recognise that ONLY the distribution of the measurements of rest mass(es) from the experiment were really specified. The distribution for relativistic mass observed was never stated and you do not have to assume the same distribution will apply. The entire experiment might only be capable of measuring rest mass anyway (if you read the description in the early part of the question). However, the relativistic masses can be calculated from the rest mass. So you can make progress because you can apply the usual rules for the propagation of errors to determine the distribution and especially the s.d. for the relativistic mass that will arise from applying that formula to a Gaussian variable. This, of course, will burn time and also give the student a different value for the standard deviation that they will end up taking forward to do the remainder of the question.
There's more I might comment on later.
Best Wishes.
Post by: Eternal Student on 21/01/2022 17:22:06
I might be talking to myself here, so I'll make this the last post unless someone else replies.
What do you think that section (g) is actually asking you to calculate? To check you've understood the question and calcuated the right thing, let's just ask if you could usefully advise a group of researchers that they should take N individual measurments if they want to "detect" the change in mass due to velocity. To phrase this another way, how you will you interpret what it means for the change in mass to be detected? Do any of the following apply?
(a) If a group of researchers take N individual measurments and average these, they will be able to determine that the relativistic mass (of protons at 0.2c) is not the same as the rest mass at the 95% confidence level. If they took less than N individual measurements then they won't be able to determine this.
(b) The results that a group of researchers obtain are random. They might be able to reject the null hypotesis (at the 95% level) that the relativistic mass is the same as the rest mass, they might not. The more individual measurments they take and combine to generate an average, the better their chances of being able to reject that hypothesis. Since everything is random they might never be able to reject the null hypothesis even if N is a massive number. However, provided they take N individual measurements and consider the mean value they obtain and the distribution of that mean in the usual way, the probability of them being able to reject that null hypothesis is 95%.
(c) 95% of the time, the mean of N individual measurements will be greater than the rest mass. The researchers should just accept this as an indication that the masses are different. They do not need to do any statistical analysis of their own, they should just go home satisified that the change in mass has been "detected".
(d) Taking N individual measurments and averaging these, the true underlying relativistic mass can be estimated with error bars as usual. We know the observed mean of N individual measurments will be Normally distributed and the standard error of the mean is easily calculated. So 95% of the time the true underlying relativistic mass will be in the range specified by (the observed mean) +/- 2 standard errors of the mean. While 5% of the time it isn't. The value of N doesn't influence these chances at all. The only thing N does is to reduce the standard error of the mean, so that the range of values reported for the relativistic mass is smaller.
(e) None of the above.
Post by: alancalverd on 21/01/2022 21:46:43
We have formulae (which I forgot about 60 years ago) for deriving confidence limits from N and σ, so we can determine the number of repeats N required to estimate A with 95% confidence.
What is not clear is whether we are required to estimate the relativistic shift from the "known true" rest mass, or from the first experimental result, which has the given standard deviation and thus influences the confidence interval of the measured change.
With my experimentalist hat on, I'm not sure I could use the same experiment to determine rest mass and relativistic mass anyway.
Post by: Halc on 21/01/2022 22:15:21
Some measurable has a true value of X and measurements have this gaussian distribution, and that's all one needs to know to answer abc.
But then the middle comes along and redefines M as relativistic mass, which hardly seems a good candidate for 'true mass'. The mathematics person doing this exercise might not know this.
So we get to g and we're torn between using 'true mass' M as per the intro, or relativistic mass (M) from the intermission. Still, the question is obviously asking for the higher mass at speed, and thus asks how many measurements should be needed to peg this new higher mass with the confidence level specified. It's still a stats problem, not a physics one. I don't see a different way to interpret g.
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Eventually you might decide that the experiment can be applied to determine relativistic mass directly AND that the distribution and standard deviation of the measurements will be unchanged.They didn't give new distributions, so you've no choice but to stick with the ones provided for the first case.
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Another reasonable course of action the student could take is to use the relativistic mass formula given earlier in the question and/or the simplified first order series they were asked to find.Then you're not 'detecting' the mass change, but simply computing it, something you can't do if you're testing the formula. The question asks for a count of samples needed in a stat exercise, and does not ask for application of physics.
BTW, I personally would fail part e because the curve is essentially flat for small values, and 0.2c (from f) isn't a small value.
What do you think that section (g) is actually asking you to calculate?I have a bag of objects with some property that yields X on average, with the distribution initially specified. I have a 2nd bag where the average is higher, but otherwise has a similar distribution.
How many samples do I need to take from each to have 95% confidence that the bags are not the same.
I think that's what it is asking.
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(a) If a group of researchers take N individual measurments and average these, they will be able to determine that the relativistic mass (of protons at 0.2c) is not the same as the rest mass at the 95% confidence level. If they took less than N individual measurements then they won't be able to determine this.Pretty much that, yes.
Post by: alancalverd on 21/01/2022 22:32:32
How many samples do I need to take from each to have 95% confidence that the bags are not the same.But the question is ambiguous as written. Do they mean the difference between the two experimental values or between the measured value Mv and the given "true" value M0 ?
Post by: Eternal Student on 21/01/2022 22:43:50
Being a simple soul, I see that the experiment produces a standard deviation of 0.2 in the determination of a number whose true value is 1.7, so the same experiment would be expected to produce σ = 0.2 A/1.7 if the true value were A.So you're basically going to scale up the standard deviation for measurments of the relativistic mass. Since the relativistic mass is 1.02 times the rest mass, you'd expect the standard deviation to be 1.02 times the original.
The good news is that you're right, Alancalverd. You'd get the same conlcusion if you used the simple (first order approxiation) for the relativistic mass that they asked the students to find earlier in the question. They would have obtained
M = M0 . (constant)
where M = relativistic mass ; M0 = rest mass and the (constant) =
. Setting v = 0.2c this would have reduced to M = 1.02 M0.
Then assuming that only the measurements of M0 were possible and had the distribution described, they would have used the conventional formula to compute the variance of a function of a known random variable.
Specifically Var( 1.02 M0 ) = 1.022 . Var (M0). Taking roots to return to standard deviations, this shows exactly the same as you (Alancalverd) had suggested.
Physicists often describe this as the "propogation of errors" which is why I used the phrase. In general, when X is a random variable and Y = aX + b is a linear function of that random variable then Y itself is a random variable and it has standard deviation and mean which is easily determined: σY= a . σX and μY = a . μX + b.
The bad news is that having seen the mark scheme, I'm afraid you would have lost one mark, Alancalverd. They did not want you to scale the standard deviation. Presumably whoever set the question, they just wanted the students to make the assumption that relativistic mass can be determined directly by the experiment and it has the same distribution (Normal and σ = 0.2 units) as for measuremenets of the rest mass.
What is not clear is whether we are required to estimate the relativistic shift from the "known true" rest mass, or from the first experimental result, which has the given standard deviation and thus influences the confidence interval of the measured change.I think I can see what you're saying. I was worried about something like that myself. Do we take as fact that M0 = 1.7 units or are we meant to assume that some group of researchers are going to try and determine if there is a mass change from scratch: They would have to get measurements for M0 first (so they don't have a single value, just a range and a 95% chance M0 is in that range) and then do more experimental measurments to get something similar (a range of values) for the relativistic mass of protons travelling at 0.2c. This makes a lot of difference since the uncertainities are almost doubled.
Anyway, looking at the mark scheme, you would have lost a mark if you had done this, along with some time that you wouldn't have to spare. You are expected to take as fact that M0 = 1.7 units.
I'm going to post this, since more replies are coming in fast - let's get something out in reply ASAP.
Best Wishes.
Post by: alancalverd on 21/01/2022 23:11:05
Presumably whoever set the question, they just wanted the students to make the assumption that relativistic mass can be determined directly by the experiment and it has the same distribution (Normal and σ = 0.2 units) as for measuremenets of the rest mass.Ugh!
Relativistic mass has a single value, given by the mass/velocity equation. The experimental problem is that a practical proton beam won't be truly monoenergetic and there are umpteen sources of random fluctuation in the measurement even if it were so, hence you will get a distribution of values for any repeated measurement. σ is a function of the method, not the phenomenon.
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(Normal and σ = 0.2 units)indeed, but the unit is now Mv, not M0 nor the experimental value M*0.
So not only is the question ambiguous,but the model answer is wrong in either case!
Just one more reason why Heisenberg's postulate should be called indeterminacy (of the quantity) to distinguish it from uncertainty (of the measurement) and must be taught as such.
Post by: Eternal Student on 21/01/2022 23:40:58
The exercise seems to be a pure mathematical one, but is confusingly worded.Yes and yes.
The title of this thread is "statistics for Physics". That exam question was taken from a course called "Quantitative Physics" and it is all about statistics.
The wording is one of the main problems. It really isn't clear at all what they want you to calculate and there are several different values of N that could be calculated. (I'll discuss more of that later).
So we get to g and we're torn between using 'true mass' M as per the intro........ etc. ...Absolutely.
Halc then talked about making the assumption that the distribution for measurments of relativisitc mass have the same characteristics....
They didn't give new distributions, so you've no choice but to stick with the ones provided for the first case.There actually is a choice. Some of this has been discussed in the earlier reply to Alancalverd.
We can assume that measurement of rest mass has that sort of distribution, we can determine the distribution that will be observed in the relativistic mass by application of what physicists often call "the propogation of errors".
You (Halc) mentioned that this is a stats. question and so there's no reason to use any physics. However I KNOW that this statistics course did cover the propogation of errors because I've helped this other person with problems based on that. The earleir parts of the question set the students up with a simple linear function to determine the relativistic mass in terms of the rest mass. They could certainly apply what they know on that.
Case in point ---> Alancalverd was going to assume the relativistic mass would show more spread.
-------
I'm going to need to take a short detour here and describe what "the experiment" is from a purely statistical point of view. As you (Halc) said, there's no need to drag much understanding of physics into this exam question:
The experiment involves a box. Into this box you put the protons. After a moment a display screen on the box shows the mass.
If you were measuring rest mass, then all you have to do is pour in protons that were stationary (relative to the box). If you were measuring the mass of protons at 0.2c then you do one of the following things:
(i) Run fast with the protons toward the box, chuck them in.
(ii) Hold the protons in something clever like a cyclotron and keep them moving inisde that container. Pour that into the box.
Basically we don't care how the box works or how we get moving protons with a definite velocity into the box. All we need to assume is that somehow the box will display a mass after protons (moving or stationary) have been put in it.
The only thing we really do need to know is what this digital display on the box will show us. Is it the relativistic mass, or is it always the rest mass? If it's the rest mass, that's OK, we can still use this. For the protons that were travelling at 0.2c we do have a simple formula connecting their relativistic mass to their rest mass, just like we have a simple formula to convert inches to centimetres. We just apply the formula and we can effectively get a set of observed values for the (relativistic) mass of protons travelling at 0.2c.
I hope this makes some sense, it's perfectly fine if "the experiment" only reports rest mass, it's just like having an instrument that reports in inches when we wanted centimetres. We can reasonably make the assumption that the box is such that there is always a little random variation in the measurments and that's the only source of random variation we are going to get. This thing, the set of mass values obtained, is what has the distribution described in the question (Gaussian with s.d. = 0.2 units).
The relativistic mass we would observe (actually it's calculated not directly observed but you would still call it an "observed measurment" because it was obtained from an observation in an experiment) has a slightly bigger standard deviation - just because we had to do some conversion. In exactly the same way that if a standard deviation was listed as +/- 1 inch then we fully expect that to become +/- 2.5 cm
- - - - - - -
Halc went on to pick answer (a) from a list of options about what the calculated value of N will tell us.
Actually, that's not right Halc. Looking at the mark scheme, you have definitely lost a mark.
Anyway, it's a valiant effort and I'm sure a lot of students would have wasted their time thinking that N can be found in this way.
This is already a long post.... I'll post now and deal with some other issues later.
Best Wishes.
Post by: Eternal Student on 22/01/2022 00:43:48
Well let's see why option (a) is incorrect. We're going to need some diagrams....
Here's the probability distribution for the relativistic mass the researchers will observe from "the experiment":
normal.jpg (38.25 kB . 1258x648 - viewed 2741 times)It's a Normal distribution. The only debateable issue is the spread, or value of the standard deviation, let's just go with what the mark scheme suggested. It is exactly the same as for observations on rest mass, we have σ = 0.2 x 10 -27 Kg.
The centre or modal point of that distribution is 1.02 times the rest mass. We know all of this from earlier parts of the question. The researchers are presumably armed with this information.
Now this is actually a lot of information we have. In particular, we don't have to take any observations to estimate the mean or standard deviation, we KNOW all of this exactly. We know about taking a mean of N values and we know all about standard errors in the mean.
Now if the researchers are lucky then they take just one measurment and it's precisely 2 sigma over to the right of the modal point of that distribution. They calculate the standard error of the mean (which for 1 observation is just exactly the same as σ). Now they know everything about this distribution, so they can write down a range of values for the true value of the relativistic mass. Specifically it is within 2 sigma of the value they just observed with 95% confidence. Well, that's nice because they observed 1.02 M0 + 2σ , so their range of values terminates at precisely 1.02 M0 and extends upto 1.02 M0 + 4σ. In particluar the lowest value is 1.02 M0 which is clearly greater than 1 M0.
So the researchers can now officially declare that the rest mass falls outside of the 95% confidence interval for the relativistic mass. That's great, they're all done, they can go home.
Anyway, you see that it is possible (maybe unlikely but still possible) that the researchers can officially detect the change in mass with just one measurement.
Conversely, if they were unlucky then they could keep getting values from over to the left side of the Normal distribution for the relativistic mass they observe. Let's say the measurments they obtain are always precisely M0. Now it doesn't matter how large N is, they can keep reducing the size of the error of the mean but it's no help. M0 would always be in the range of values they will state for their 95% confidence interval just because their range of values is centred on M0. So it is possible (unlikely but still possible) that even if N was 1 million, they might still not be able to officially declare that the rest mass lies outside the 95% confidence interval for the relativistic mass.
Anyway, option (a) is definitely false. There is no certainty that the researchers can rule out the possibility that M and M0 are the same at the 95% confidence level even when N is very large. Also, there is some chance to rule it out even when the researchers take considerably less than N observations.
Post by: alancalverd on 22/01/2022 15:44:32
Case in point ---> Alancalverd was going to assume the relativistic mass would show more spread.No!
The experimental determination of relativistic mass Mv is likely to show a significant spread, and the given M0 "true rest mass" had no attached σ, so the best first guess is that σv, experimental ≈ σ0, experimental. However as I pointed out, you can't measure M0 and Mv by the same method, so even that assumption is invalid.
Nor does the question actually mention testing a null hypothesis between two experiments. It simply asks for the number of trials needed to establish Mv = 1.02 M0,true (or maybe M0,expt) with a 95% confidence interval. Or perhaps it implies a null hypothesis where σ (M0,expt ) ≠ 0.
How ambiguous can you get without actually being a politician?
Eddington said "If you need statistics, you should have done a better experiment".
In my experience mathematicians are harmless if kept in their cages and fed with diffusion constants, but can cause untold confusion if let loose in a laboratory. I had a theoretical physicist colleague who for the purposes of libel laws I shall call Smith. When we built a new experimental rig we fitted a handwheel with a big "DO NOT ADJUST" notice and an axle running through the wall to a pointer labelled "Smith Detector" in the office next door. Excellent correlation was observed. He retired as a professor of theoretical physics, his wife having ensured that he never owned a soldering iron.
Post by: Eternal Student on 22/01/2022 16:35:32
I'm very grateful to everyone who has spent some time on this thread.
It is my opinion that the question has these issues:
1. Poor wording in general. [with some agreement from Halc and Alancalverd]
2. Use of the word "mass" in a way that is ambigious - since it could be taken to mean "rest mass" only. [agreement from Halc]
3. Lack of clarity about the distribution characteristics for whatever this "mass" measurment will be [agreement from Alancalverd, who would have scaled up the s.d. ]
4. Lack of clarity on whether the rest mass can be taken as fact or if it would also have to be determined by experiment. [agreement from Alancalverd].
5. None of us demonstrated an appreciation of what that number, N, would actually do. [Halc chose option (a) from an earlier post (which is impossible, you would never find a value of N that did that, hopefully Halc would have moved on to the next question after 10 minutes), I spent a lot of time considering something that would fit option (b) when I first did the question - that is possible to find but time consuming, it gives a value approximately 4 x bigger than they wanted but more importantly it's not what you were supposed to calculate. Alancalverd never guessed what N would do but grumbled about how the situation was unreaslistc].
Having seen the mark scheme (and with the benefit of far more time than would have been available in the exam), it is my opinion that the criteria they (whoever set that question) consider appropriate for detecting the change in mass is actually quite arbitrary.
They consider the change in mass to be "detectable" if it is twice the standard error for the mean. While this might sound superficially sensible, as previously discussed in earlier threads this will not guarantee that any group of researchers would detect that change at the 95% confidence when they do their own statistics.
- - - - -
A different view
The requirement that the change in mass is twice the standard error for the mean, is, at best, a nominal criteria for being able to detect the change in mass.
I need to be fair and present some argument why this criteria is nominally sensible. Here's one argument:
1. Consider a large number of observations, N, being made. So that N observed relativistic mass measurments are avalable.
2. We can reasonably expect the observed values to cluster around 1.02 M0.
3. So any group of researchers should have an observed mean approximately equal to 1.02 M0 (if N is large).
4. The researchers can compute the standard error for the mean as usual, σmean = σ / √N.
5. So we can reasonably expect them to report the relativistic mass as being within 2. σmean of 1.02 M0.
6. Provided M0 is further away from 1.02 M0 than this, then it won't be in that range . The researchers will be able to declare that the relativistic mass is different to the rest mass at the 95% confidence level.
So that's it. A fair attempt to justify the criteria they have used for detectability. It's a nominal criteria and nothing more. It's more justified when the calculated value of N does turn out to be large.
Best Wishes.
Alan just wrote something....
No! The experimental determination of relativistic mass Mv is likely to show a significant spread....Yes, that's just me not putting too many extra words in. I did see your earlier post. We appreciate that the relativistic mass is a fact, it is only the experimental determination of the relativistic mass that could show some spread.
(... Alancalverd has edited his own post since I started this ... I'm just going to post this or else we'll never get anything finished and posted).
Post by: Eternal Student on 22/01/2022 16:46:05
Nor does the question actually mention testing a null hypothesis between two experiments. It simply asks for the number of trials needed to establish Mv = 1.02 M0,true (or maybe M0,expt) with a 95% confidence interval.Well that's actually very good and close enough to what they were asking you to calculate (see earlier post, if you're interested).
Or perhaps it implies a null hypothesis where σ (M0,expt ) ≠ 0. ....How ambiguous can you get without actually being a politician?I'll take that as an indication that you would also have lost some time trying to make sense of the question. (I certainly did and it seems Halc would have done the same). I think we're all in agreement that this is not a well crafted question.
Best Wishes.
Post by: Eternal Student on 22/01/2022 17:04:44
Attached is the mark scheme (almost model answer) that was circulated by that University to this years students.
I know, Halc suggested avoiding just printing an answer but the extenuating circumstances seem strong enough to me. The model answer is already "out there" and we have thoroughly discussed plenty of issues. Any of their students accidentally finding this thread have a fair opportunity to learn something if they tried reading the thread.
To be honest it would take them longer to read the thread than just to try the question.
Moderators ---> You are welcome to remove or edit this post without consulting with me, if you do not agree.
Best Wishes.
Post by: Eternal Student on 23/01/2022 22:06:17
@alancalverd
Basically thank you very much for your time and expertise.
I would have been talking to myself otherwise.
I'm a lot happier now and reasonably re-assured that the advice I gave to this other person wasn't entirely silly. That was an extended family member by the way, whch is not important except that there was never any commercial element to the whole thing.
Anyway, you helped an old person (me) and also you helped to educate a person new to the science scene. You did well.
Thanks and best wishes.