0
$\begingroup$

I've been thinking about special relativity and I did the following math (I’m a beginner in this relativity so sorry if the problem I did has mistakes)

Let’s suppose a spaceship travelling at $\frac{1}{2}$ light speed ($150.000 km/h$) in a vacuum measures a photon that travels $300.000 km$ in one second. If we are still and we apply special relativity to what the spaceship has measured in comparison with us we find the folowing

The $300.000 km$ for the spaceship is $346.410,16 km$ for us($300.000 x \gamma(150.000)$), and the $1$ second for the spaceship is $1,155 s$ for us ($1 x \gamma(150.000)$). If we divide the results to get the speed ($346.410,16/1,155$) we get $300.000 km/h$ , the speed of light, so we see that it’s constant and doesn’t vary no matter at what speed we measure it.

The problem I found is that if I change those numbers for others, for example the speed we're measuring is (measured by the spaceship) $ 250.000 km/h$ , that speed stills the same: $250.000 km$ for the spaceship is $288.675,13 km$ for us. $1 s$ for the spaceship is $1,155 s$ for us $\frac{288.675,13}{1,155 }= 250.000 km/h$

So $250.000 km/h$ is behaving the same as light speed, being constant

Can someone explain why is this happening to me?

Improve this question
$\endgroup$
6
  • 1
    $\begingroup$ Hint: Pick two events A and B, at which the photon is present. (In your own coordinates, you might as well call A the event where t=0, x=0 and B the event where t=1,x=1). Now compute the coordinates of those events in the frame of the spaceship. Divide to get the speed of light in the spaceship's frame. Now pick two events A and C at which a particle traveling (relative to you) at 5/6 the speed of light is present. You might as well call A the event where t=0,x=0 and C the event where t=1, x=5/6. Convert to the spaceship's frame and divide to get a velocity. Is it still 5/6? $\endgroup$
    – WillO
    Apr 20, 2018 at 21:29
  • $\begingroup$ I still get 5/6, I don’t know where I messed up. Please help me, not having an awnser for this is very stressing. $\endgroup$
    – Espartex200
    Apr 21, 2018 at 23:38
  • $\begingroup$ How are you still getting $5/6$? What did you get for the coordinates of $A$ and $C$ in the spaceship's frame? $\endgroup$
    – WillO
    Apr 22, 2018 at 2:32
  • $\begingroup$ (Incidentally: Consider the extreme case where the particle travels at speed $0$ with respect to the ground-based observer. Then clearly it is traveling at a speed of $-1/2$ relative to the spaceship, right? So if your calculation tells you that all speeds are invariant, try applying your calculation to the case of speed zero and see where it goes wrong.) $\endgroup$
    – WillO
    Apr 22, 2018 at 2:40
  • $\begingroup$ I can’t believe that I made such a stupid mistake, THANKS. That really helped ÷). $\endgroup$
    – Espartex200
    Apr 22, 2018 at 10:08

2 Answers 2

Reset to default
1
$\begingroup$

First we need to clarify:

  1. your calculations seem to be correct.

  2. you are setting the speed of light to 250000 km/s

  3. then your calculations seem to be correct too.

  4. you are wondering why the calculation still works with a different speed of light.

  5. the Maxwell equations set the speed of light to approx. 300000 km/s in the universe. It is because the vacuum is set up that way, the permittivity and permeability of vacuum define that EM waves can propagate through vacuum with this exact speed.

  6. you wonder why this is the case, but you just have to accept that EM waves, photons and every particle without rest mass can travel with this speed, because the vacuum is set up so in the universe, vacuum has EM fields everywhere, and photons are just an excitation in the field. Now this excitation is what propagates with this speed. We do not know why, but maybe because EM fields need time to excite and so the propagation of this excitation needs time.

  7. what you are doing is you are trying to set this characteristic of the universe to a different value, in your case to 250000 km/s. That would need a different vacuum permittivity and permeability.

  8. In this case, EM waves would need a different time to propagate through vacuum.

  9. This would still not change SR, so any observer would see light propagate at this 250000 km/s speed.

Improve this answer
$\endgroup$
0
$\begingroup$

SR (special relativity) is the outcome of the two Einstein principles: 1) Each inertial reference frame is equivalent, 2) The speed of light is a constant in every inertial reference frame; plus assumptions on the homogeneity and isotropy of vacuum.
The first principle alone yields that in any inertial reference frame there is a limit speed, which is a constant independent of the frame and that can not be overcome by any material object. According to experimental evidence the speed of light fits perfectly this requirement. That is why of the second principle.
Of course, if you manipulate the Lorentz transformations and instead of $c$ you post another limit speed $c_l$ formally the formulas work, but you depart from a correct description of the reality, as it is measured.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.