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Acceleration vs Gravity

  1. Aug 2, 2015 #1
    Dear PF Forum,
    In Einstein elevator, Einstein states that we can't tell between gravity and acceleration.
    I have incoming questions concerning gravity, that's why I create a different thread.
    Can we use this formula for gravity?
    Case A:
    A rest observer watches an accelerating object and that object keeps sending signal containing its clock. The rest observer compares its clock to the object's clock with the above formula, is this true?
    Case B:
    What if there is a rest observer, "outside" the Earth gravitational field. And supposed the Earth is floating in space, no near planet no sun nearby by millions of years. The observer and the earth distance is always the same (I know, the observer has to orbit the earth no matter how really slow it moves. The gravitational fields extend infinitely before it is disrupted by other celestial object, just supposed the speed of the object revolves the earth is very small).
    Does the observer watches the clock on the earth by that formula above as the observer watches an accelerated object 9.8m/sec2?
  2. jcsd
  3. Aug 2, 2015 #2


    Staff: Mentor

    No, that's not what he states. He states that we can't tell being held at rest in a gravitational field from acceleration. For example, if you are in a small enclosed room with no windows, feeling your normal weight of 1 g, you can't tell, just from measurements inside the room, whether it is sitting at rest on the surface of the Earth or out in deep space somewhere inside a rocket whose engine is providing 1 g of thrust. But in both cases, the force you feel, that causes you to feel weight, is not gravity; it's whatever is pushing on you (the Earth's surface or the rocket's engine).

    No. The two cases you describe are different, because in case A, the accelerating object's distance from the rest observer changes, while in case B, it doesn't. The formula you give only applies to case A.
  4. Aug 2, 2015 #3
    Yes, of course! The distance changes! In case A the object is (keeps) changing his frame of reference. Or at least the object is in a different frame than the observer, even with out accelerating, just velocity only. How can I didn't think of that. I'll contemplate your answer. Thanks.
  5. Aug 2, 2015 #4


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    The acceleration in this case comes from some propulsion engine. The formula is from the coordinate chart in flat ( no gravity) spacetime that describes an object with continuous constant acceleration.
    Yes, I think that is right. ##T## is the clock rate of the accelerating object compared with that of the rest object.
    This is gravitational time dilation. It has a different formula, which depends on gravitational potential, not acceleration. See Pound & Rebka experiment.

  6. Aug 2, 2015 #5


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    Gold Member

    1. The given formula does not give the instantaneous difference in clock rates at any given moment "t". It gives the comparative accumulated time. In other words, if a clock has a proper acceleration of 'a' for a time 't' as measured in the non-accelerated frame, then a time of T will have accumulated on the accelerated clock. The instantaneous clock rate difference is always changing, so the accumulated time T will not just be a simple multiple of the clock rate at that moment.

    To get the instantaneous clock rate you first find the instantaneous velocity of the the clock with respect to the non-accelerated frame. This is

    [tex]v = \frac{at}{\sqrt{1+\frac{a^2t^2}{c^2}}}[/tex]

    Then you plug this 'v' into the SR time dilation equation. When you do this you end up with:

    [tex] \frac{1}{\sqrt{1-\frac{1}{1+\frac{c^2}{a^2t^2}}}}[/tex]

    Which is the instantaneous time rate factor between the accelerated and non-accelerated clock. assuming that when t=0, the velocity difference between the clocks was 0.

    2. Even this second formula doesn't give you a gravitational time dilation equivalent. Gravitational time dilation is not due to the difference in g forces acting on the clocks but due to the difference in gravitational potential. For a acceleration/gravity equivalence explanation consider the explanation I gave with your accelerating train scenario. Remember that according to the observers in the accelerating train, they saw a Doppler shift between the front and back of the train even though, in their frame the distance didn't change, and thus they would conclude that the clock in the front end of the train runs fast compared to the one at that rear end. This would be true even the front and rear of the train felt the same proper acceleration.

    Now consider light traveling upwards from the surface of the Earth. As it climbs, it loses energy fighting gravity, however unlike everyday objects, it doesn't lose speed. It give up energy by changing frequency. So the observer above the surface sees this light as red shifted. Like above, this observer cannot attribute any of this red shift to an increasing distance between the surface of the Earth and himself, so it is entirely attributed to time dilation. Thus for this observer, time runs slower at the surface than it does for him. For the surface observer measuring light coming from above him, it gains energy and he sees a blue shift, causing him to to conclude that time in regions above him runs faster. This is like the observers in the train noting the difference between the clock rate difference between the front and back.
  7. Aug 2, 2015 #6
    Actually Mentz114 has pointed out that the accelaration differs in space coordinate than gravity. That's the one that I should have realized.
    I didn't ask this. But I didn't realize that there is clock rate, either. Thanks anyway for giving me this answer. Now I know that there is clock rate.
    Yes, thanks.
    A couple of time you gave me this answer in my previous post. But still struggling with the basic SR, the non accelerated one. Perhaps I'll give it time to analyze.
    Wow, I didn't realize that! Which I should!. All this time I live with the idea that the speed of light is always, ever and ever invariant. And in my unconsious mind I must have tought that light climbs the gravity well loses speed.
    Okay what about in black hole?
    1. Does the frequency of the light is zero MHz, instead of Orange, 500 THZ, or Green, 600 Thz?
    2. Does the light ever comes out at all with zero hertz frequency?
    Last edited: Aug 2, 2015
  8. Aug 3, 2015 #7
    Can I change the equation into this?
    ##V = \frac{at}{\sqrt{1+a^2t^2}}##?

    Can I change that into this?
    Last edited: Aug 3, 2015
  9. Aug 3, 2015 #8
    Yes, provided you measure time in metres. Also remember that this formula only applies for constant proper acceleration cases.
  10. Aug 3, 2015 #9
    You mean "provided you measure v in a factor of c"? I suspected that much, but I need confirmation. I'm afraid if I were wrong it could mislead me to understand the rest of the post.
    Ahh, sorry. It's c. It is the speed of light.
    I think we can't do that. It's not like ##1+\frac{v}{c} \rightarrow 1+v##, right?
    Btw, still reading your link. It's a very, very good link. But I'm trying to make sense this thread first.
  11. Aug 3, 2015 #10
    Yes, and we do that by meauring time in meters, where 1 meter of time is the amount of time it takes for light to travel 1 meter of distance. When we do this, velocities becomes dimensionless (the meters cancel out) and c becomes 1.
  12. Aug 3, 2015 #11
    But I realize there's no V in this equation.
    ##V = \frac{at}{\sqrt{1+\frac{a^2t^2}{c^2}}}## Do you really think we can convert it to ##V = \frac{at}{\sqrt{1+a^2t^2}}##
    Supposed, what is V at 10 sec
    ##V = \frac{at}{\sqrt{1+\frac{a^2 10^2}{c^2}}} \rightarrow \frac{at}{\sqrt{1+a^2 (3,000,000)^2}}##?Is this true?
  13. Aug 3, 2015 #12
    No, you divide a by 3 million squared. You use the c in the formula if you're using seconds.
  14. Aug 3, 2015 #13
    Okay, no c elimination. Just use a and t unchanged.
  15. Aug 3, 2015 #14
    I think this is what I didn't know that I didn't know.
    We often eliminate c in this equation.
    ##S = \frac{u+v}{1+\frac{uv}{c^2}} \rightarrow S = \frac{u+v}{1+uv}## where u and v is in a factor of c.
    ##\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} \rightarrow \gamma = \frac{1}{\sqrt{1-v^2}}## where v is a factor of c.
    When I created this thread:
    I couldn't express my question. Now I KNOW what I don't know. Perhaps we should leave a and t as it is. No c elimination.
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