Why Is It Impossible To Reach The Speed Of Light?

[novop]

From the second line, you're using the time dilation result which is a consequence of the invariance of the speed of light in all inertial frames, and is indeed predicated on v < c. Still, it's nice to arrive at the same result coming from different paths.

 

[stevmg]

Can anyone show me if sin^-1(v/c) = a₀t/c is a monotonic increasing function of t with an upper limit on v/c (or $$\beta$$) of 1, hopefully where this range on v/c is a half open set [0, 1)?

 

[stevmg]

From the second line, you're using the time dilation result which is a consequence of the invariance of the speed of light in all inertial frames, and is indeed predicated on v < c. Still, it's nice to arrive at the same result coming from different paths.

novop, my brain works in different paths each time and I never can reproduce what I have done before.

That's correct and just try and squeeze in the v $$\geq$$ c... All of the equations fall down - like a house of cards.

 

[diazona]

Can anyone show me if sin^-1(v/c) = a₀t/c is a monotonic increasing function of t with an upper limit on v/c (or $$\beta$$) of 1, hopefully where this range on v/c is a half open set [0, 1)?

The domain of the arcsin function (the set of values of v/c for which the function is defined, which I assume is what you mean) is [-1,1]. That's just the set of values that you can get by taking the sine of a number.

But arcsin(x) is monotonically increasing, as long as you restrict it to the range $-\pi/2 \le x \le \pi/2$. Just plot it on a graph to see that.

 

[stevmg]

The domain of the arcsin function (the set of values of v/c for which the function is defined, which I assume is what you mean) is [-1,1]. But it is monotonically increasing. Just plot it on a graph and you'll see that.

You're correct, diazona, I got my domain and range turned around.

Instead of "But arcsin(x) is monotonically increasing, as long as you restrict it to the range $-\pi/2 \le x \lx \pi/2$." you meant "$-\pi/2 \le + \lx \pi/2$," right?

 

[diazona]

I meant $-\pi/2 \le x \le \pi/2$ (now fixed). Just a little typo

 

[stevmg]

I meant $-\pi/2 \le x \le \pi/2$ (now fixed). Just a little typo

Yea, but my damn equation is wrong... back to the drawing board on that one.

 

[stevmg]

Regarding the recurrent statements that, in SR, it always assumed that v is less than c.

Actually, when I look at Einstein's derivation of the Lorentz transformations in Relativity, 15th edition, 1952, Appendix I, he makes no assumption on v. At least I cannot see any. As he derives the equations for x' and t' he comes up with that infamous
$\gamma$^-1 = $\sqrt{(1 - v^2/c^2)}$
That square root puts the kibosh on any v > c in the real world.

Please correct me, as I am probably wrong and have missed something.

 

[Mentz114]

Regarding the recurrent statements that, in SR, it always assumed that v is less than c.

Actually, when I look at Einstein's derivation of the Lorentz transformations in Relativity, 15th edition, 1952, Appendix I, he makes no assumption on v. At least I cannot see any. As he derives the equations for x' and t' he comes up with that infamous
$\gamma$^-1 = $\sqrt{(1 - v^2/c^2)}$
That square root puts the kibosh on any v > c in the real world.

Please correct me, as I am probably wrong and have missed something.

That's correct. When v = c, $\gamma$ 'blows up', and if v > c the square root is imaginary.

 

[RK1992]

This is almost a valid argument apart from a technical error in the maths. You have treated $\gamma$ as a constant, but it isn't. I will leave it as an exercise for the reader that the correct formula turns out to be

$$\frac{dv}{dt}=\frac{F}{\gamma^3\,m}$$​

But the final conclusion still holds true.

I get that $\gamma$ is really a function in v so you have some type of differential equation where velocity is your variable but I don't know how you got the ^3 in there..

 

[stevmg]

I get that $\gamma$ is really a function in v so you have some type of differential equation where velocity is your variable but I don't know how you got the ^3 in there..

Yea... I agree with RF1992, where does $\gamma^3$ come from?

Makes sense that Fdt = $\gamma$mdv (a variation of the relativistic momentum equations as referred to by starthaus) where m is the mass you start off with.

 

[stevmg]

Relevant to the above... can anyone out there "integrate" or solve this diff eq?

a₀ is a given value (say, 9.8 m/sec²) at t = 0

dv/dt = a₀$\sqrt({c^2 - v^2/c^2})$

dv/$\sqrt({c^2 - v^2})$ = a₀dt/c

sin^-1 (v/c) = a₀t/c

Now, something I have done is wrong... what?

 

[DrGreg]

This is almost a valid argument apart from a technical error in the maths. You have treated $\gamma$ as a constant, but it isn't. I will leave it as an exercise for the reader that the correct formula turns out to be

$$\frac{dv}{dt}=\frac{F}{\gamma^3\,m}$$​

But the final conclusion still holds true.

I get that $\gamma$ is really a function in v so you have some type of differential equation where velocity is your variable but I don't know how you got the ^3 in there..

The neatest way to show this is to make the substitution $v = c \tanh \phi$, so that $\gamma = \cosh \phi$, so

$$F = \frac{dp}{dt} = \frac{d}{dt} \left( mc\,\sinh \phi \right) = mc \, \cosh \phi \, \frac{d\phi}{dt}$$​

and

$$\frac{dv}{dt} = \frac{d}{dt} \left(c \tanh \phi \right) = c \, \mbox{sech}^2 \phi \, \frac{d\phi}{dt}$$​

Divide one by the other.

If you're not comfortable with hyperbolic functions, begin with

$$F = \frac{dp}{dt} = \frac{d}{dt} \left( \gamma m v \right) = m \left( \frac{d\gamma}{dv}\, \frac{dv}{dt} \, v + \gamma \frac{dv}{dt} \right)$$​

and calculate $d\gamma/dv$. A bit more painful, but you'll get there in the end.

 

[stevmg]

The neatest way to show this is to make the substitution $v = c \tanh \phi$, so that $\gamma = \cosh \phi$, so

$$F = \frac{dp}{dt} = \frac{d}{dt} \left( mc\,\sinh \phi \right) = mc \, \cosh \phi \, \frac{d\phi}{dt}$$​

and

$$\frac{dv}{dt} = \frac{d}{dt} \left(c \tanh \phi \right) = c \, \mbox{sech}^2 \phi \, \frac{d\phi}{dt}$$​

Divide one by the other.

If you're not comfortable with hyperbolic functions, begin with

$$F = \frac{dp}{dt} = \frac{d}{dt} \left( \gamma m v \right) = m \left( \frac{d\gamma}{dv}\, \frac{dv}{dt} \, v + \gamma \frac{dv}{dt} \right)$$​

and calculate $d\gamma/dv$. A bit more painful, but you'll get there in the end.

Hey... that's weird but it's right. OMG! Not very inuitive but right.

Mustn't forget that basic rule of calculus: if w and t are functions of x, then d(wt)/dx = wdt/dx + tdw/dx.

 

[RK1992]

The neatest way to show this is to make the substitution $v = c \tanh \phi$, so that $\gamma = \cosh \phi$, so

$$F = \frac{dp}{dt} = \frac{d}{dt} \left( mc\,\sinh \phi \right) = mc \, \cosh \phi \, \frac{d\phi}{dt}$$​

and

$$\frac{dv}{dt} = \frac{d}{dt} \left(c \tanh \phi \right) = c \, \mbox{sech}^2 \phi \, \frac{d\phi}{dt}$$​

Divide one by the other.

If you're not comfortable with hyperbolic functions, begin with

$$F = \frac{dp}{dt} = \frac{d}{dt} \left( \gamma m v \right) = m \left( \frac{d\gamma}{dv}\, \frac{dv}{dt} \, v + \gamma \frac{dv}{dt} \right)$$​

and calculate $d\gamma/dv$. A bit more painful, but you'll get there in the end.

Okay so I can do it now, thanks :) I have no idea about hyperbolic functions (age 17, not studied them yet) so I went for the ugly approach and I can see it.

How on Earth, though, do we show that:

$$\frac{v^{2}}{\sqrt{1-v^{2}/c^{2}}^{3}} + \frac{1}{\sqrt{1-v^{2}/c^{2}}} = \frac{1}{\sqrt{1-v^{2}/c^{2}}}^{3}$$

I believe it because I graphed it but how do we do the maths of it? thanks you've already been incredibly helpful, I love this forum :D

 

[diazona]

You're missing something in there somewhere, the units don't match up. Maybe should it be v²/c² in the numerator of the first fraction, instead of just v²?

Anyway, once you do that, it's just like adding any other fractions. Convert them to a common denominator and add the numerators.

 

[RK1992]

You're missing something in there somewhere, the units don't match up. Maybe should it be v²/c² in the numerator of the first fraction, instead of just v²?

You're right.. I'll give it a go now I've corrected it

Edit: yep it was simple, lol..it's an elegant way of showing the acceleration decreases, I like it.

 

[Fredrik]

I'm using units such that c=1. If you want to restore factors of c, just replace every v in my calculations with v/c.

Your equation says that [itex]v^2\gamma^3+\gamma=\gamma^3[/tex]. This is equivalent to $v^2\gamma^2+1=\gamma^2$, and we have

$$v^2\gamma^2+1=\frac{v^2}{1-v^2}+\frac{1-v^2}{1-v^2}=\frac{1}{1-v^2}=\gamma^2$$

By the way, I found the lack of parentheses in your expression confusing at first. You can get parentheses of different sizes with \big(, \Big(, \bigg(, \Bigg(, or \left(. The last one has to be closed with \right).

Edit: By the way, the calculations I did here might be interesting (as an exercise, if nothing else).