# Relativity mass question **URL:** \ **Category:** Factual Questions\ **Created:** [March 17, 2016, 12:25am UTC](https://boards.straightdope.com/t/relativity-mass-question/749221 "2016-03-17T00:25:41Z")\ **Posts on this page:** 4\ **Page:** 2
**Author:** ![Chronos](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/chronos/32/134_2.png) [@Chronos](https://boards.straightdope.com/u/Chronos)\ **Post date:** [March 18, 2016, 7:48pm UTC](https://boards.straightdope.com/t/relativity-mass-question/749221/21 "2016-03-18T19:48:55Z")
One does sometimes still use the term “rest mass” just to be completely unambiguous, but when dealing with others in the field who know what is meant, the “rest” is unnecessary. Given a choice of ways to phrase something, scientists tend towards the less ambiguous way. ---
**Author:** ![EastUmpqua](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/eastumpqua/32/6025_2.png) [@EastUmpqua](https://boards.straightdope.com/u/EastUmpqua)\ **Post date:** [March 18, 2016, 10:21pm UTC](https://boards.straightdope.com/t/relativity-mass-question/749221/22 "2016-03-18T22:21:56Z")
> [@Chronos](#): > > One does sometimes still use the term “rest mass” just to be completely unambiguous, but when dealing with others in the field who know what is meant, the “rest” is unnecessary. Given a choice of ways to phrase something, scientists tend towards the less ambiguous way. Didn’t you say that backwards? ---
**Author:** ![Chronos](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/chronos/32/134_2.png) [@Chronos](https://boards.straightdope.com/u/Chronos)\ **Post date:** [March 19, 2016, 1:28pm UTC](https://boards.straightdope.com/t/relativity-mass-question/749221/23 "2016-03-19T13:28:05Z")
I don’t think so. Scientists don’t like ambiguity, so we try to decrease it, and so sometimes end up adding extra words that wouldn’t be strictly speaking necessary, like the “rest” in “rest mass”. ---
**Author:** ![Asympotically\_fat](https://avatars.discourse-cdn.com/v4/letter/a/e47c2d/32.png) [@Asympotically\_fat](https://boards.straightdope.com/u/Asympotically_fat)\ **Post date:** [March 19, 2016, 7:16pm UTC](https://boards.straightdope.com/t/relativity-mass-question/749221/24 "2016-03-19T19:16:45Z")
> [@Senegoid](#): > > And, the difficulty of accelerating a massive object is, in some sense, the very definition of what mass is. Mass is sometimes defined (or explained) simply as _that_ property, possessed by physical objects, of being “difficult” (that is, requiring force) to accelerate. In Newtonian physics you have: **f** = m **a** Or in other words force (a vector) equals mass (a scalar) times acceleration (a vector). The inertial mass is a very straightforward measure of how much force it takes to produce a certain acceleration in an object, or in other words how much an object resists a change in its state of motion In special relativity sometimes the so-called relativistic mass is sometimes presented as the equivalent of inertial mass, but force is not generally equal to relativistic mass times acceleration. I thought it would be fun (for very small values of fun) to see exactly what you have to multiply by the acceleration to get the force that produces it in special relativistic context, as I’ve never seen this explicitly done before and what I get is: **f** = **M\*\*\*\*a** **M** can’t be a scalar like rest mass or relativistic mass, as in special relativity an acceleration is not always parallel to the force that produces it. **M** is actually the 3x3 matrix: ```auto (m[sub]0[/sub]γ[sup]3[/sup]v[sub]x[/sub][sup]2[/sup]/c[sup]2[/sup] + m[sub]0[/sub]γ m[sub]0[/sub]γ[sup]3[/sup]v[sub]x[/sub]v[sub]y[/sub]/c[sup]2[/sup] m[sub]0[/sub]γ[sup]3[/sup]v[sub]x[/sub]v[sub]z[/sub]/c[sup]2[/sup] ) (m[sub]0[/sub]γ[sup]3[/sup]v[sub]y[/sub]v[sub]x[/sub]/c[sup]2[/sup] m[sub]0[/sub]γ[sup]3[/sup]v[sub]y[/sub][sup]2[/sup]/c[sup]2[/sup] + m[sub]0[/sub]γ m[sub]0[/sub]γ[sup]3[/sup]v[sub]y[/sub]v[sub]z[/sub]/c[sup]2[/sup] ) (m[sub]0[/sub]γ[sup]3[/sup]v[sub]z[/sub]v[sub]x[/sub]/c[sup]2[/sup] m[sub]0[/sub]γ[sup]3[/sup]v[sub]z[/sub]v[sub]y[/sub]/c[sup]2[/sup] m[sub]0[/sub]γ[sup]3[/sup]v[sub]z[/sub][sup]2[/sup]/c[sup]2[/sup] + m[sub]0[/sub]γ) ``` Where m[sub]0[/sub] is the rest mass, v[sub]x[/sub], c is the speed of light in a vacuum, v[sub]y[/sub] and v[sub]z[/sub] are the components of the objects velocity along the x, y and z axes and γ is relativistic gamma which is equal to [1 - (v[sup]2[/sup]/c[sup]2[/sup])][sup]-0.5[/sup] [Previous page](https://boards.straightdope.com/t/relativity-mass-question/749221.md?page=1)