[alancalverd]

m_v =  m₀(1 - v²/c²)^−1/2

This works for "relativistic speeds" but clearly raises problems when v = c, as Einstein remarked. And who am I to disagree? 

[PmbPhy]

m_v =  m₀(1 - v²/c²)^−1/2

This works for "relativistic speeds" but clearly raises problems when v = c, as Einstein remarked. And who am I to disagree?

That didn't help me understand your sentence, i.e. what did you mean when you wrote "We're not talking about "relativistic speeds" but c."?

[alancalverd]

Pretty much the same as the difference between "green-ish" and "λ = 520 nm". 0.99c is a relativistic speed in most people's language. c is c.

The point is that all and only photons travel at c, whereas any particle with nonzero mass could in principle travel at any speed less than c. That makes c rather special, and 0.999c rather ordinary.

[lightarrow]

Pretty much the same as the difference between "green-ish" and "λ = 520 nm". 0.99c is a relativistic speed in most people's language. c is c.

c is not relativistic speed? I don't know if laughing or crying...

The point is that all and only photons travel at c, whereas any particle with nonzero mass could in principle travel at any speed less than c. That makes c rather special, and 0.999c rather ordinary.

Your problem is the obstination to want to use  equations which are valid only for velocities different from c. If you used those which are always valid, for all velocities included c, you wouldn't need to "climb on the mirrors"  []

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lightarrow

[evan_au]

what happens when an object follows a circular path at light speed?

We're not talking about "relativistic speeds" but c.

what did you mean when you wrote "We're not talking about "relativistic speeds" but c."?

I think we may be talking at cross purposes here.

The OP was asking about swinging an object around at lightspeed=c.
The discussion moved on to discussing what happens at relativistic speeds < c.

Alan was just reminding us that the OP was asking about traveling at speed=c, which is infinitely harder than traveling at "relativistic speeds" < c...

[alancalverd]

Your problem is the obstination to want to use  equations which are valid only for velocities different from c. If you used those which are always valid, for all velocities included c, you wouldn't need to "climb on the mirrors"  []

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lightarrow

I don't have a problem! I'm trying to answer the original question. By all means choose any equation you like that consists with experimental evidence, and tell us how much energy we need to expend to accelerate an object for which m₀≠ 0 to the speed of light, then how much force is required to constrain it to a circular path with a constant tangential speed c.

[lightarrow]

Tangential Radial (centripetal) force. In modulus: F = (E/r)*β²

E = energy.
β = v/c

For a body with mass: E = mc²γ; γ = (1-β²)^-1/2
For a body moving at c: β = 1.

About the energy needed to accelerate a body to light speed, see "E" up. For a photon you can write E = h*f if you like.

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[jeffreyH]

If we take the value we end up with . How do you come to this determination? Beta may equal 1 according to your previous post but then you are taking the square root of zero and dividing it into 1. How is that valid? None of this has a bearing upon c.

[PmbPhy]

Tangential force. In modulus: F = (E/r)*β²

E = energy.
β = v/c

For a body with mass: E = mc²γ; γ = (1-β²)^-1/2
For a body moving at c: β = 1.

About the energy needed to accelerate a body to light speed, see "E" up. For a photon you can write E = h*f if you like.

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lightarrow

Did you know that in his 1905 paper on special relativity Einstein got the expression for the transverse mass wrong? Ohanian explains the error in his book Einstein's Mistakes located at http://www.newenglandphysics.org/other/Ohanian_Einstein_error_1905.pdf
http://www.newenglandphysics.org/other/Ohanian_Einstein_error_1905.pdf

[lightarrow]

If we take the value we end up with . How do you come to this determination? Beta may equal 1 according to your previous post but then you are taking the square root of zero and dividing it into 1. How is that valid? None of this has a bearing upon c.

Sorry, I'm a bit "slow" in these days, what do you mean exactly? Are you referring to the equation F = m*a when F is in the same direction of velocity?

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[alancalverd]

Tangential force. In modulus: F = (E/r)*β²

E = energy.
β = v/c

For a body with mass: E = mc²γ; γ = (1-β²)^-1/2
For a body moving at c: β = 1.

Agreed, so as v→c, E → ∞ if m ≠ 0

And the radial force required to maintain circular motion of a massive body at v = c is?

[lightarrow]

Tangential force. In modulus: F = (E/r)*β²

E = energy.
β = v/c

For a body with mass: E = mc²γ; γ = (1-β²)^-1/2
For a body moving at c: β = 1.

Agreed, so as v→c, E → ∞ if m ≠ 0

And the radial force required to maintain circular motion of a massive body at v = c is?

Ehm, I wrongly wrote "tangential" force but it clearly was "radial" force. Sorry for the mistake and thank you to have noticed it!
I am correcting it.

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lightarrow

[lightarrow]

Tangential force. In modulus: F = (E/r)*β²

E = energy.
β = v/c

For a body with mass: E = mc²γ; γ = (1-β²)^-1/2
For a body moving at c: β = 1.

About the energy needed to accelerate a body to light speed, see "E" up. For a photon you can write E = h*f if you like.

--
lightarrow

Did you know that in his 1905 paper on special relativity Einstein got the expression for the transverse mass wrong? Ohanian explains the error in his book Einstein's Mistakes located at http://www.newenglandphysics.org/other/Ohanian_Einstein_error_1905.pdf
http://www.newenglandphysics.org/other/Ohanian_Einstein_error_1905.pdf

Very interesting.
So, after all, Albert Einstein wasn't that "super mind" being we are used to think, but just a common person which makes the same kinds of mistakes you and me can make too  []

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[alancalverd]

Ehm, I wrongly wrote "tangential" force but it clearly was "radial" force. Sorry for the mistake and thank you to have noticed it!
I am correcting it.

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lightarrow

Correction noted, and as I stated way back in this thread, F → ∞ if v → c and m ≠ 0. Glad we settled that.