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E=mc^2 no object with rest mass can ever reach the speed of light?

  1. May 17, 2005 #1
    How does [tex]E = mc^2 [/tex] imply that no object with rest mass can ever reach the speed of light?
  2. jcsd
  3. May 17, 2005 #2
  4. May 17, 2005 #3

    The equation for the total energy in special relativity is the same, but the mass in question is γMo, where γ is velocity-dependent and γ is 1 when velocity is zero (respect to the observer) and tends to infinity when velocity tends to c (light velocity). Mo is the restmass (respect to an observer moving with the object. So you would need an infinite amount of energy to get the speed of light.
  5. May 20, 2005 #4
    anyone correct me if im wrong, but isnt it because when a particle approaches the speed of light its mass increases towards infinity, and when it hits the speed of light its mass is infinity, so according to E=mc^2 you would need an infinite amount of energy, which u cant get
  6. May 20, 2005 #5


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  7. May 20, 2005 #6
    ah, now I understand :D thx everyone

  8. May 20, 2005 #7
    But this would also mean that the "m" variable in e=mc^2 can be relavistic mass right?
  9. May 20, 2005 #8
    yes, it is.
  10. May 20, 2005 #9


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    Yes - AFAIK the only time E is not equal to Mc^2 is the case Pete likes to bring up that occurs when a system is not isolated (i.e. when there are forces acting on it from the "outside world").

    For an isolated system, the concept of the relativistic mass of a system can be replaced by the concept of the energy of the system.

    One might argue that no system is truly isolated. But as far as the magnitude of the effect goes, the critical point is when the energy of interaction of the system with the outside world becomes a significant fraction of the total energy of the system. "Significant" depends a lot on the experimental accuracy that one desires.
  11. May 23, 2005 #10
    Ok i have a question thats a little off topic but its been on my mind for a long time and would much appreciate an anwser, here it goes. According to past forums i have read relatavistic mass does not create warps in space-time, in other words energy does not create warps in space-time, but instead it is rest mass that effects gravity, now in a nuclear reaction it is rest mass that is used as m in E=mc^2 right? and if this is right then that means that rest mass equals energy, so energy would effect gravity also right?
  12. May 23, 2005 #11


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    Energy, momentum, and pressure all cause gravity via the "stress-energy tensor".

    I'm not sure where you read that relativistic mass does not warp space-time. The phrase is unclear enough that I'm not sure quite what it means.

    Some examples might (or might not) help clarify the situation.

    Suppose you have a cold object. From an external source, you heat it up. The cold object now has a greater gravitational field. It also has a greater invariant mass. The two are closely related by the equivalence principle - so gravitatioanl mass will always be equal to inertial mass, or just invariant mass.

    Suppose you have a massive object, and you either a) increase it's velocity or b) increase your velocity relative to it. The gravitational field of the moving object is different from that of the non-moving ones in many ways (it's not spatially uniform, for one thing), but a rapidly moving object will never (for instance) turn into a black hole.

    The last point to consider is a combination of the first two points above. Suppose we have a gas in a container, and we heat the gas up while keeping it contained.

    Then the difference between the hot container and the cold container is just the fact that the gas molecules are moving around, in all directions.

    We've already argued that the mass (and gravity) of the system with the hot gas is (very slightly!) greater than the mass of the system with the cold gass. As long as the gravity is "weak-field", we can decompse the total gravitational field of the system into the sum of the fields due to each particle. Thus we can argue that in some sense the field of an individual particle is "stronger" when it is moving, but this is an average sort of field over many directions, since the field of a moving particle is definitely not uniform.

    This approach eventually breaks down - eventually, the entire idea of gravity "as a field" starts to have problems. This occurs when the space-warping effects of gravity start to become important. At this point, a uniform treatment that treats gravity as a curvature of space-time becomes necessary. To fully explain the math of this concept is difficult, about all I can say is that it is important to know that the idea of gravity as a "field" eventually gets replaced by the notion of curvature.
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    Last edited: May 23, 2005
  13. May 24, 2005 #12
    I am begining to question everything I know about physics. I have knows several physisics, and things I hear from them and physicsforums.com often contradict eachother. This case in particular. I am not critisizing what you wrote, I am just saying some things you said contradict other things I have heard.

    I will just have to learn it myself in colledge :tongue:
  14. May 24, 2005 #13


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    Well, I can provide a few online references for what I say. The only other thing I can suggest is bringing some of the people who say different things than we say online so that we can argue with them :-).

    The sci.physics.faq does a pretty good job of explaining why a fast-moving object doesn't turn into a black hole.


    (It doesn't give any references for further reading, though - a lot of the sci.physics.faq do, but this particular one doesn't).

    This leaves the question of whether a hot object is more massive than a cold one.


    is about the only non-technical reference (and you'll see a certain amount of confusion in the responses, only the response by Dick Plano really confirms the point that a hot object has a higher mass).

    On the same topic

    http://scitation.aip.org/getabs/servlet/GetabsServlet?prog=normal&id=AJPIAS000066000005000409000001&idtype=cvips&gifs=yes [Broken]
    also online in the arivx at

    is rather technical, but much more explicit in giving an affirmative answer.

    There are not any really good article on the gravitational field of a relativistically moving body. There are certain problems with the very notion of a gravitational field when relativistic velocities are considered. One could measure the gravitational field of a moving body by forcing an accelerometer to move "in a straight line" with a constant velocity near a massive object. The readings of the accelerometer forced to move in this manner would give the "gravitational field" of the moving object. The problem with this idea is that space is curved near a massive body, and the defintion of what a "straight line is" becomes ambiguous. At small velocities, this isn't a big issue, but at relativistic velocities it is.
    Last edited by a moderator: May 2, 2017
  15. May 25, 2005 #14
    i was talking about relativistic mass does not effect gravity and has no gravitational force
  16. May 25, 2005 #15


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    I think you may have misunderstood what other members on physicsforums.com said, I remember you made a similar accusation of people contradicting each other on this thread but in that case you were misunderstanding, and nothing said on that thread contradicts what pervect said above. Can you point to a specific post that you think does contradict it?
  17. May 31, 2005 #16
    Hi pervect. You're one sharp dude and I consider myself honored that you referred to my previous discussions. :approve:

  18. May 31, 2005 #17
    I miss these arguements. Perhaps I'll get my ISP back today.

    "Relativistic mass" sure does effect gravity. Claims otherwise are extremely incorrect. This fact can be found anywhere. Even in MTW's text. I think I've even given examples with all the math and everything. I want to make a clearer example soon so stay tuned.

    More - Its important to keep this fact in mind. From "The Philosophy of Space and Time," Hans Reichenbach, page 214
    Its important to keep in mind what the term "covariant" meas in this quote. It refers to the components of tensors and how they change upon change in coordinates. Quantities/numbers which have this property are called "covariant" while those which don't change are called "invariant." There are many different uses of the term "covariant" so best to keep them distinct.

    Last edited: May 31, 2005
  19. May 31, 2005 #18
    Its rarely the case that a body does not interact with the outside world. Consider a uniform E-field. Place a dumbell in the field and its mass won't be related to the energy by E = mc^2. The dumbell being a rod with a -q charge on one end and +q charge on the other end. The rod won't accelerate. It will move at constant velocity in all inertial frames. But the mass won't read as E = mc^2.

    Its odd that even Rindler missed this fact in a paper he wrote. He got his own definition of mass wrong in an example!

    Consider also other complex examples; let a box which is initially at rest in S be of finite size. Let the faces be perpendicular to the axes. Let the box emit radiation of equal amounts in opposite directions parallel to the x-direction. Now boost in the x-direction and observer the box emitting radiation. The relation [itex]E = \gamma m_0 c^2[/itex] will fail to work for all time even though the box is isolated. Wanna guess why? :biggrin:

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