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Author Topic: Is e=mc^2 the same as F=ma^2 ?  (Read 9271 times)

Offline Thebox

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Is e=mc^2 the same as F=ma^2 ?
« on: 09/09/2015 17:26:32 »
 E=mc˛ is exactly the same as F=ma˛ when (c) and (a) are accelerations of the same magnitude?
« Last Edit: 11/09/2015 08:14:26 by chris »


 

Offline alancalverd

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Re: A question of relationship?
« Reply #1 on: 09/09/2015 18:20:33 »
Except that F = ma, not ma2

F is a vector, not a scalar,

c is a speed, not an acceleration.

In other words, no, not at all.
 

Offline Thebox

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Re: A question of relationship?
« Reply #2 on: 09/09/2015 18:47:15 »
Except that F = ma, not ma2

F is a vector, not a scalar,

c is a speed, not an acceleration.

In other words, no, not at all.


I did not ask  any of the above Alan , I asked

 E=mc˛ is exactly the same as F=ma˛ when (c) and (a) are accelerations of the same magnitude?


If m=m  and c=a would this not give the same answer?

and curiosity, I now want to know what is the acceleration of light if you are saying c is only a speed?
« Last Edit: 09/09/2015 18:49:02 by Thebox »
 

Online PmbPhy

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Re: A question of relationship?
« Reply #3 on: 09/09/2015 18:55:16 »
Quote from: Thebox li
I did not ask  any of the above Alan , I asked

 E=mc˛ is exactly the same as F=ma˛ when (c) and (a) are accelerations of the same magnitude?
No. Those aren't the same. In fact F = ma^2 is dimensionally incorrect and not an equality.
 

Offline Thebox

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Re: A question of relationship?
« Reply #4 on: 09/09/2015 18:59:18 »
Quote from: Thebox li
I did not ask  any of the above Alan , I asked

 E=mc˛ is exactly the same as F=ma˛ when (c) and (a) are accelerations of the same magnitude?
No. Those aren't the same. In fact F = ma^2 is dimensionally incorrect and not an equality.

F=m1*a

F=M2*a

F=ma˛

does this not describe a collision?

If not sorry my misunderstanding.

 

Offline jeffreyH

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Re: A question of relationship?
« Reply #5 on: 09/09/2015 21:20:26 »
An acceleration is metres per second squared. Whereas c is in meters per second. Acceleration is a derivative of velocity. Is shows how the velocity changes over time. Velocity is a vector and is directional whereas speed isn't. It can be in any direction. Thus the speed of light is not a velocity. The useful thing about velocity is that it relates to energy. E=mc^2 can be thought of as the rest energy. For motion velocity can be used to derive the kinetic energy, or energy of motion. The equation is 1/2mv^2 where v is always less than c. I hope this helps.
 

Offline alancalverd

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Re: A question of relationship?
« Reply #6 on: 09/09/2015 23:34:18 »
I now want to know what is the acceleration of light if you are saying c is only a speed?

Light doesn't accelerate. c is a universal constant.

Obviously you can write a = bc2 and x = yz2, and if b = y and c = z, then a = x. But the symbols you used have particular conventional meanings.
 

Offline jeffreyH

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Re: A question of relationship?
« Reply #7 on: 09/09/2015 23:44:49 »
Take a simple situation where a = b + c. Say a = 10, b = 4 and c = 6. We can square both sides of the equation as in a^2 = (b + c)^2. To expand (b + c)^2 we first have (b + c)(b + c). This multiplied out gives b^2 + 2bc + c^2. Working this out we get a^2 = 10^2 which is 10 times 10 so that is 100. Now we have to find a value of 100 from b^2 + 2bc + c^2. b^2 is 4^2 which equals 16. c^2 is 6^2 which equals 36. So from those we get 16 + 36 which equals 52. Now we get to 2bc which is 2 times 4 times 6. So two times 4 is 8 and 6 times 8 = 48. Then adding 52 to 48 hey presto we get 100. Squaring values on both sides of an equation can be very useful when solving some types of equation.
 

Offline jeffreyH

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Re: A question of relationship?
« Reply #8 on: 10/09/2015 00:10:11 »
As a good first step in advancing your understanding of probability a well known real life example is useful.

https://en.wikipedia.org/wiki/Lottery_mathematics
 

Offline Thebox

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Re: A question of relationship?
« Reply #9 on: 10/09/2015 16:43:39 »





Hi Alan, taking you back to E=mc˛   , I thought this was a velocity ?   and from what I understand a Photon smashing into the side of a virtual box and moving the box?

how is that not the same as force if that is correct?

 

Offline Thebox

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Re: A question of relationship?
« Reply #10 on: 10/09/2015 16:44:36 »
Take a simple situation where a = b + c. Say a = 10, b = 4 and c = 6. We can square both sides of the equation as in a^2 = (b + c)^2. To expand (b + c)^2 we first have (b + c)(b + c). This multiplied out gives b^2 + 2bc + c^2. Working this out we get a^2 = 10^2 which is 10 times 10 so that is 100. Now we have to find a value of 100 from b^2 + 2bc + c^2. b^2 is 4^2 which equals 16. c^2 is 6^2 which equals 36. So from those we get 16 + 36 which equals 52. Now we get to 2bc which is 2 times 4 times 6. So two times 4 is 8 and 6 times 8 = 48. Then adding 52 to 48 hey presto we get 100. Squaring values on both sides of an equation can be very useful when solving some types of equation.

Thank you Jeffrey , can you please explain E=mc˛ to me, and post an example of this works?
 

Offline Thebox

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Re: A question of relationship?
« Reply #11 on: 10/09/2015 16:46:09 »
As a good first step in advancing your understanding of probability a well known real life example is useful.

https://en.wikipedia.org/wiki/Lottery_mathematics

Isn't there now 59 balls?  I do not play or watch the lottery but I heard they added an extra ten balls
 

Online PmbPhy

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Re: A question of relationship?
« Reply #12 on: 11/09/2015 10:30:34 »
Quote from: Thebox
F=m1*a

F=M2*a

F=ma˛

does this not describe a collision?

If not sorry my misunderstanding.
It's not possible to scratch down equations without stating their meaning and expect it to represent what's going on. I.e. you can't simply write down equations and ask what's going on in the lab. In fact all you've done here is scratch out equations without stating what the variables mean. If we assume that F = force and m, m1, M2 represent the mass of something then F=ma˛ is dimensionally incorrect and therefore wrong. E.g. Force has dimensions

[F] = Newton's = [kg][m/s2]

The right side of your equation has dimensions

[kg][m2/s4]

So dimension wise your last equation reads

[kg][m/s2] = [kg][m2/s4]

which is obviously incorrect. You have to keep in mind that you can't just write down expressions and expect them to mean something. If you thought that's what physics was all about then you were seriously mistaken.

By the way "c" is speed whereas "a" is acceleration. Why did you think that the "c" in E = mc2 was acceleration?
« Last Edit: 11/09/2015 11:09:37 by PmbPhy »
 

Offline mathew_orman

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Re: Is e=mc^2 the same as F=ma^2 ?
« Reply #13 on: 11/09/2015 11:02:36 »
E=mc˛ is exactly the same as F=ma˛ when (c) and (a) are accelerations of the same magnitude?
No, c is a constant and a is a variable...
Also, one models energy and the other one, force...
 

Offline Thebox

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Re: A question of relationship?
« Reply #14 on: 11/09/2015 19:44:32 »


By the way "c" is speed whereas "a" is acceleration. Why did you think that the "c" in E = mc2 was acceleration?

The speed a photon travels from the instant of emittance is c?


So from a to b , light travels at instantly c, c being the speed and acceleration?

and

[kg][m/s2] →F← [kg][m2/s4]

a collision?



 

Offline alancalverd

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Re: Is e=mc^2 the same as F=ma^2 ?
« Reply #15 on: 11/09/2015 22:09:49 »
No. c is a speed, not a velocity nor an acceleration.The speed of light does not vary in a vacuum. No variation = no acceleration.

E = mc2 has nothing to do with photons colliding with the walls of a box. It is the energy released when masses annihilate, or the mass of particles produced when a photon interacts with a nucleus, or a whole lot of other stuff that Pete will go on about, bit nothing to do with momentum transfer.
 

Offline Thebox

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Re: Is e=mc^2 the same as F=ma^2 ?
« Reply #16 on: 12/09/2015 08:30:27 »
No. c is a speed, not a velocity nor an acceleration.The speed of light does not vary in a vacuum. No variation = no acceleration.

E = mc2 has nothing to do with photons colliding with the walls of a box. It is the energy released when masses annihilate, or the mass of particles produced when a photon interacts with a nucleus, or a whole lot of other stuff that Pete will go on about, bit nothing to do with momentum transfer.

I should stress  to people not to watch youtube videos of science , the thought experiment to Einstein's E=mc˛ shows a photon hitting an imaginary box. You mention annihilation of mass, do you mean like a Neutron star I think it was, that crushes protons?


I do not understand why E=mc˛ has a speed attached to it, or uses massless light when you say it is mass related.


Why would Energy be related to a speed and not a compression?


''The speed of light does not vary in a vacuum. No variation = no acceleration.''


So are you saying that if I had a vacuum tunnel that was 299 792 458 meters long, light would take one second to travel the distance and this never alters?
A clock constant that does not alter to a said time dilation like the materialistic values of the Caesium atom shows a frequency offset by gravity influence.
« Last Edit: 12/09/2015 08:32:40 by Thebox »
 

Offline jeffreyH

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Re: Is e=mc^2 the same as F=ma^2 ?
« Reply #17 on: 12/09/2015 12:53:06 »
No. c is a speed, not a velocity nor an acceleration.The speed of light does not vary in a vacuum. No variation = no acceleration.

E = mc2 has nothing to do with photons colliding with the walls of a box. It is the energy released when masses annihilate, or the mass of particles produced when a photon interacts with a nucleus, or a whole lot of other stuff that Pete will go on about, bit nothing to do with momentum transfer.

I should stress  to people not to watch youtube videos of science , the thought experiment to Einstein's E=mc˛ shows a photon hitting an imaginary box. You mention annihilation of mass, do you mean like a Neutron star I think it was, that crushes protons?

Youtube can be good and bad. I recommend finding lecture videos where they write examples on a white board and discuss what they mean. I would start simply with algebra first. Being shown something is 1000 times better than reading about it. Don't get discouraged because there will be points where you suddenly get an understanding of something. Then it starts to get interesting.

I do not understand why E=mc˛ has a speed attached to it, or uses massless light when you say it is mass related.

Momentum is mass times velocity mv. That is how far a mass moves in a particular direction in a certain period of time. Mass tends to resist change hence inertia. It either wants to stay in one place or move at a constant velocity. The direction of this velocity will be a straight line. E=mc2 relates to rest mass which since it is at rest is not moving in any direction. This is why the speed of c is used and not a velocity. Kinetic energy, the energy of motion through space, has to be directional and is given by Ke = (1/2)mv2. Here the mv2 represents motion with velocity because it is directional. The formula E=mc2 uses the speed of light simply because we are concerned with energy that is not moving and has no direction.

Why would Energy be related to a speed and not a compression?


Ask yourself what energy is. If something is completely still then the energy of motion is absent and only rest energy remains. If energy related to compression then everything at rest would be feeling that compression. Objects would tend to compact and get smaller which we do not see in the real world.

''The speed of light does not vary in a vacuum. No variation = no acceleration.''


So are you saying that if I had a vacuum tunnel that was 299 792 458 meters long, light would take one second to travel the distance and this never alters?
A clock constant that does not alter to a said time dilation like the materialistic values of the Caesium atom shows a frequency offset by gravity influence.

The statement that light travels at c through a vacuum does not take into account any forces. Once you introduce a force, or some kind of medium that light has to travel through, then this changes.
 

Offline alancalverd

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Re: Is e=mc^2 the same as F=ma^2 ?
« Reply #18 on: 12/09/2015 14:15:34 »


Why would Energy be related to a speed and not a compression?




Energy = force x distance. In the case of adiabatic compression, the kinetic energy expended in compressing a hysteresis-free body resides as potential energy.   

But since force = mass x acceleration and acceleration = distance /time^2, so energy has the dimensions of mass x distance^2/time^2

Speed has dimensions distance/time, so mc^2 has the same dimensons as energy. 
 

Offline Thebox

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Re: Is e=mc^2 the same as F=ma^2 ?
« Reply #19 on: 13/09/2015 10:18:11 »


Energy = force x distance. In the case of adiabatic compression, the kinetic energy expended in compressing a hysteresis-free body resides as potential energy.   

But since force = mass x acceleration and acceleration = distance /time^2, so energy has the dimensions of mass x distance^2/time^2

Speed has dimensions distance/time, so mc^2 has the same dimensons as energy.


Ok I think I understand what you are saying, however energy can not be associated with Photons,  photons are a product of a process, are you saying without photons energy would not exist?  if this is the case then where did the energy come from for the big bang to create things in the first place?
E=mc˛ makes no relative sense.
Energy can only be formed under a space compression such as plasma and magnetic bottling, otherwise space just ''dissolves'  the energy.
The energy will expand into space has energy has no ''physical body'' or glue, gravity is like a glue that holds matter together, this allows the matter to store energy and to contain energy.


P.s I think I might be Bi-polar so thank you for your patience.
« Last Edit: 13/09/2015 10:20:58 by Thebox »
 

Offline alancalverd

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Re: Is e=mc^2 the same as F=ma^2 ?
« Reply #20 on: 13/09/2015 10:27:57 »

E=mc˛ makes no relative sense.


Pity about that! My colleagues in the nuclear medicine department use it every day, and 20% of your electricity comes from that equation, so you will just have to accept that it is magic, or study physics with more humility and less preconception. Expect to have your common sense violated frequently.
 

Offline mathew_orman

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Re: Is e=mc^2 the same as F=ma^2 ?
« Reply #21 on: 13/09/2015 10:42:26 »
 

E=mc˛ makes no relative sense.


Only if you miss use it with photons...
It was meant to be used with particles which have properties of mass...

 
 
 

Offline Thebox

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Re: Is e=mc^2 the same as F=ma^2 ?
« Reply #22 on: 13/09/2015 10:46:55 »



Only if you miss use it with photons...
It was meant to be used with particles which have properties of mass...


I still can not understand why a speed has anything to do with energy, lightning forms with no speed and lightning is energy is it not?
 

Offline mathew_orman

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Re: Is e=mc^2 the same as F=ma^2 ?
« Reply #23 on: 13/09/2015 10:57:01 »
Because kinetic energy is only defined if there is relative motion...
 

Offline Thebox

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Re: Is e=mc^2 the same as F=ma^2 ?
« Reply #24 on: 13/09/2015 12:22:37 »
Because kinetic energy is only defined if there is relative motion...

surely energy exists with no motion required?
 

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Re: Is e=mc^2 the same as F=ma^2 ?
« Reply #24 on: 13/09/2015 12:22:37 »

 

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