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So I've recently been studying relativity a lot trying to understand it and I feel like I grasp most things conceptually but I have one issue I've been trying to understand for the last couple days and I just can't seem to find an answer anywhere.

Let's say you're traveling very fast, somewhere close to the speed of light. At this point you're in an inertial reference frame. From this frame you shouldn't be aware that you are anywhere near the speed of light. Now you want to accelerate by 5 m/s to a new reference frame that is even closer to the speed of light. Is there any difference in the amount of energy required to accelerate to the new reference frame in this scenario from the amount required to accelerate by 5 m/s from a slower reference frame like here on Earth?

I sort of understand the concept of relativistic energy but if that somehow applies here I don't get how. If you are in an inertial reference frame, then it seems that the amount of energy required to accelerate to a new reference frame should be the same no matter what your frame is. However, from various things I've read I get the impression that the energy required to accelerate at a constant rate is not constant, which seems to make sense from a static reference from but not from an accelerating one.

I'm sure I'm missing something or there's a flaw in my thinking somewhere. I hope this all makes sense since I'm still very new to this. Also, if there's a way to explain the concept with little to no math that would be very helpful since I still don't grasp a lot of the math involved in relativity. I'll take what I can get though. This has been eating at me too much.

Update: For anyone wondering about this same thing I finally found the answer explained in a way I grasped it here.

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  • $\begingroup$ What are you traveling relative to? Your velocity is observer dependent. It's not a relevant physical quantity unless you have two systems. Neither is this a problem of special relativity. You have exactly the same situation in Galilean relativity. What your kinetic energy is and by how much it increases or decreased if you accelerate is observer dependent. How much energy you need to accelerate depends on where the accelerating force comes from. If it is exerted by something that moves really fast relative to you, it will take more energy than a force that comes from a slow moving object. $\endgroup$ – CuriousOne May 26 '16 at 5:40
  • $\begingroup$ When you say, "How much energy you need to accelerate depends on where the accelerating force comes from" I think that is the part I'm not understanding. So what if you're in a rocket of some sort and you fire it? What if the force is coming from within your reference frame or one close to it? $\endgroup$ – Justin Warkentin May 26 '16 at 5:57
  • $\begingroup$ A rocket is not one reference frame but at least two. It's the payload and the mass the rocket expels. If you remember Newton, a rocket, as a whole doesn't really move. The center of mass always stays at the launch point. This is a perfectly classical problem and you need to go back to the derivation of kinetic energy again. Do you remember why it goes with the square of the velocity? $\endgroup$ – CuriousOne May 26 '16 at 6:02
  • $\begingroup$ In your (initial) rest frame suppose your speed increases to $dv$ then your energy change is $\sqrt{p^2c^2+m^2c^4} - mc^2$, where $p=\gamma mdv$. In my frame your velocity increased from $v$ to $v'$, where you have to calculate $v'$ using the relativistic addition of velocity ($v'\ne v+dv$). Then in my frame your energy change is $\sqrt{p'^2c^2+m^2c^4} - \sqrt{p^2c^2+m^2c^4}$, where $p'$ and $p$ are the initial and final relativistic momentum. I'll leave you to do the calculation since it's messy, unilluminating and would contravene our homework policy. $\endgroup$ – John Rennie May 26 '16 at 6:38
  • $\begingroup$ Ok, I think this is making sense. I also just found another good discussion of the same thing here. It helps knowing the right thing to search for. Thanks guys! $\endgroup$ – Justin Warkentin May 26 '16 at 6:53
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In special relativity the transition from one frame to another is given by the Lorentz boosts. This is not quite the same as an acceleration, but a transformation that relates observations on one frame with another. We can think of an acceleration as being a succession of infinitesimal Lorentz boosts that map one frame to another.

The infinitesimal distance in flat spacetime $ds$ for a particle is given by $$ ds^2 = c^2dt^2 - dx^2 - dy^2 - dx^2 $$ where this distance $ds – cd\tau$, for $\tau$ the time as measured by a clock on the frame of the particle. Let us consider the motion of this particle in the $x$ direction. Now divide through by $ds^2$ to get $$ 1 = \left(\frac{dt}{d\tau}\right)^2 - \frac{1}{c^2}\left(\frac{dx}{d\tau}\right)^2. $$ We can write this according to four-velocity $U_t = \frac{dt}{d\tau}$, $U_x = \frac{dx}{d\tau}$ $$ 1 = U_t^2 - U_x^2. $$ Now take the derivative of this with respect to $\tau$ so that $$ 0 = \left(\frac{dU_t}{d\tau}\right)U_t - \frac{1}{c^2}\left(\frac{dU_x}{d\tau}\right)U_x. $$ This leads to the interesting observation that in spacetime the four-acceleration is perpendicular to the four-velocity.

This system of equations leads to a solution for the four-velocity $$ U_t = cosh(g\tau),~U_x = c~sinh(g\tau), $$ for $g$ the acceleration. We can see that the equation defines a hyperbola in $t, x$ coordinates. For large $\tau$ that the hyperbola is approximately $U_t^2 = U_x^2$, and it is not hard to get these in the $t, x$ coordinates. We can also see that the coordinate based velocity is $$ \frac{dx}{dt} = \frac{U_x}{U_t} = c~tanh(gt), $$ which indicates this particle asymptotes to the speed of light as $\tau\rightarrow\infty$.

When it comes to energy we appeal to the four-momentum interval in special relativity $$ m^2 = E^2 - p^2 $$ where $E = mU_t$ and $p = mU_x$ are energy and spatial momentum respectively. Using the properties of hyperbolic trigonometric functions this can be seen. We can see right off that energy is given by a hyperbolic cosine function that diverges enormously as $\tau\rightarrow\infty$.

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