Relativistic length contraction

[Sriman Dutta]

Hi friends,

 

Once again I got this situation in my head.

Let there be a metal rod of original length l, and the rod material has a coefficient of linear thermal expansion of a. Let this rod travel towards a star at a speed v, very near to the speed of light. Let the change in temperature of the rod be T. Then new length after thermal expansion l' = l(1+aT). But there is also length contraction. So the real length l" = l'/y = {l(1+aT)}/y , where y is the Lorentz factor.

Am I right in this calculation ? Or, am I missing out some other factors ?

[DimaMazin]

Hi friends,

 

Once again I got this situation in my head.

Let there be a metal rod of original length l, and the rod material has a coefficient of linear thermal expansion of a. Let this rod travel towards a star at a speed v, very near to the speed of light. Let the change in temperature of the rod be T. Then new length after thermal expansion l' = l(1+aT). But there is also length contraction. So the real length l" = l'/y = {l(1+aT)}/y , where y is the Lorentz factor.

Am I right in this calculation ? Or, am I missing out some other factors ?

Don't complicate relativity by classical physics.

[studiot]

Don't complicate relativity by classical physics.

 

 

 

+1 You don't get many of these, but you deserve one here.

[swansont]

Don't complicate relativity by classical physics.

Relativity is a classical theory.

Hi friends,

 

Once again I got this situation in my head.

Let there be a metal rod of original length l, and the rod material has a coefficient of linear thermal expansion of a. Let this rod travel towards a star at a speed v, very near to the speed of light. Let the change in temperature of the rod be T. Then new length after thermal expansion l' = l(1+aT). But there is also length contraction. So the real length l" = l'/y = {l(1+aT)}/y , where y is the Lorentz factor.

Am I right in this calculation ? Or, am I missing out some other factors ?

The change in temp has to be in the rod's frame.

[Tim88]

Hi friends,

 

Once again I got this situation in my head.

Let there be a metal rod of original length l, and the rod material has a coefficient of linear thermal expansion of a. Let this rod travel towards a star at a speed v, very near to the speed of light. Let the change in temperature of the rod be T. Then new length after thermal expansion l' = l(1+aT). But there is also length contraction. So the real length l" = l'/y = {l(1+aT)}/y , where y is the Lorentz factor.

Am I right in this calculation ? Or, am I missing out some other factors ?

 

There's a beginner's mistake in phrasing: the speed v is not absolute, you may just as well pretend that the rod ends up in rest, and so the Earth is moving at speed v. Therefore you can't say that it's the "real" length!

 

Apart of that it looks OK to me. Assuming that you meant it as swansont specified, then that's the length according to Earth.

[mathematic]

The "real" length is that due to the thermal expansion. The relativistic correction is that seen by an outside observer moving at speed v relative to the rod.

[Sriman Dutta]

The original length is what I assumed the length of the rod at the beginning of the situation. Yah, it's true that the rod's state of rest may not be the same for another observer, because according to relativity there is no absolute frame of reference.

[Tim88]

Oh in fact you did forget gravitational length contraction, if you wanted to include GR.

[Sriman Dutta]

Hi Tim88, I know very little about GR as I can't seem to understand the concept of tensors in the EFE.

[Tim88]

The link sends you to a section without tensors, and the text is understandable. But yeah the symbols are difficult to figure out.

 

The equation in Wikipedia for time dilation is correct:

https://en.wikipedia.org/wiki/Gravitational_time_dilation#Outside_a_non-rotating_sphere

 

That's the time dilation factor (see also just above that for the connection with the approximate equation). The length contraction factor in the radial direction (perpendicular to the surface) is the same.

 

Note that the contraction is only in that direction; there is no gravitational length contraction in the tangential direction (parallel to the surface).