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I just had a confusion. Does the mass of the body actually increase when it is moving with a certain velocity? Or does it only look like the mass has increase to another observer. How can the actual mass of the body increase. Please correct me if I am wrong but I feel that it only seems to the observer that the mass has increased but it does not increase in real.

The increase of actual mass would imply that that its velocity will decrease to comply with conservation of momentum. But how can the velocity decrease if no external force is applied?

Qmechanic
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rahulgarg12342
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    Well the answer you are referring to does not answer my question in detail as I need the reason (why?). But thanks for your reference. – rahulgarg12342 Jan 01 '14 at 16:14

2 Answers2

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There isn't a simple answer to your problem because it depends on what you mean by mass.

You've probably heard of Einstein's equation $E = mc^2$, and if you rearrange this equation you can use it as a definition of the mass:

$$ m = \frac{E}{c^2} $$

For a stationary object the mass we get is the rest mass or invariant mass, which is what we normally mean by mass. However if the object is moving then it has some kinetic energy so:

$$ E_{total} = mc^2 + \text{kinetic energy} $$

and we can define a relativistic mass using:

$$ m_r = \frac{E_{total}}{c^2} $$

and for a moving particle this relativistic mass is obviously greater than the invariant mass. This is mathematically a possible definition for the mass, but in the 109 years since special relativity was formulated it has caused generations of students much confusion, just as it is confusing you now in 2014. To avoid this we now rarely refer to relativistic mass, and modify Einstein's equation to be:

$$ E^2 = p^2c^2 + m^2c^4 $$

where $m$ is the invariant mass and $p$ is the momentum. Though I won't duplicate it here, the Wikipedia article on Mass in special relativity gives a proof that the two equations are the same.

You ask if the mass increase is real, but that can't be answered because you have to define what you mean by real. The difference between relativistic and invariant mass comes from accounting for the energy in different ways. One is no more real than the other, though the concept of invariant mass is much less confusing.

IkelosOne
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John Rennie
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The inertia (resistance to acceleration) increases, and the object's tendency to curve space around it increases as it's velocity approaches the speed of light. But physicists have found it cumbersome to treat mass as a variable, or to keep saying "rest mass" all the time, so we say that the mass-proper stays the same.

Edit:

Some clarification in response to John Rennie's comment is in order. The gravitational field surrounding a small, massive particle, like a ball bearing, is given by the Schwarzschild metric, thus we are no longer considering purely special relativistic effects. If a test mass is moving with respect to the particle (as measured from a great distance), one would consider a boosted Schwarzschild metric.

If one is considering a distribution of mass in motion, the relevant object is the stress-energy tensor, which is the source term for the metric. The $T^{00}$ term is what you would normally call the energy of the particle, and is $\gamma m v^2 \delta(\mathbf{x}-\mathbf{x}_p(t))$ for a single moving particle. Things get a bit murky when you specify exactly what you mean by $\mathbf{x}(t)$, here, and you're best off taking $m$ to be much smaller than the other length scales in your spacetime.

lionelbrits
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  • So that means the actual mass does increase as you said that its gravity (tendency to curve more space) and inertia increase. Is my conclusion right? If it is then it will mean that its velocity will decrease to comply with conservation of momentum. But how can the velocity decrease if no external force is applied? – rahulgarg12342 Jan 01 '14 at 13:44
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    You say the object's tendency to curve space around it increases, but I don't think this is true. The metric is an invarient and looks the same from all coordinate systems. If travelling faster increased the curvature that would imply that an object moving fast enough could become a black hole. – John Rennie Jan 01 '14 at 13:47
  • I completely agree with you John and that is what I personally believe. But I need to reach a conclusion. Does it actually increase because I am writing a research paper and had this confusion. I cannot seem to get a proper answer anywhere. – rahulgarg12342 Jan 01 '14 at 13:53
  • @John: The Minkowski metric is invariant wrt Lorentz boosts, but it isn't generally true. – lionelbrits Jan 01 '14 at 15:56
  • Btw, @rahulgarg12342 , the momentum is $\gamma m\mathbf{v}$. The momentum only increases when you accelerate the object. Here $\gamma$ is a function of velocity, $\gamma = \sqrt{\frac{1}{1-\frac{v^2}{c^2}}}$. As the object approaches the speed of light, $\gamma \to \infty$. Note that $m$ is just a constant. – lionelbrits Jan 01 '14 at 17:19
  • But the metric and stress-energy tensors are tensors and therefore coordinate independent. Their components will change when you move to a boosted frame, e.g. $T_{00}$ will change as you say, but the tensor as a whole will not. – John Rennie Jan 01 '14 at 19:56
  • What is meant is that if a hyper-relativistic object flew by you then the retarded gravitational force you would feel at some point (say, closest approach) would be greater than that of a non-relativistic object of the same mass. The tensors are invariant wrt diffeomorphisms but the relative distance between you and the object as a function of time is the relevant parameter here. – lionelbrits Jan 02 '14 at 11:40