# A E = mc2 Condundrum

#### JSHD2019

Summary
Can you cancel the masses out by turning E = mc2 into a space lever?
E = mc2 : kg.m2/s2 ≡ kg.m.m/s2 ≡ kg.m.a ≡ m.N ≡ Nm ≡ j as requ'd
Thence,
Distance x F = m.C2
Distance x Acceleration = C2
For a rest mass likely only Force acting is due to Gravity so:
Distance x 9.81 = C2
Distance = 9.161e+15 meters

Can you do the above or is the reasoning non valid and illogical? Your views

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#### BvU

Homework Helper
Hello JSHD, $\qquad$ $\qquad$ !

Can't do, I 'm afraid :

• The potential energy expression for the potential energy from gravity , $mgh$ is only valid in the neighborhood of the surface of the earth ($g$ has to be constant).

• $E=mc^2$ is in fact $E=\gamma m_0 c^2$ where $m_0$ is the rest mass (-- the one you use for $mgh$ )

And a few more issues ...

But keep the questions coming !

$\$

#### russ_watters

Mentor
For a rest mass likely only Force acting is due to Gravity so:
Distance x 9.81 = C2

Can you do the above or is the reasoning non valid and illogical?
You can't do that.

#### JSHD2019

ok if not a lever how about a wavelength so e.g.

1 light year = 9.461e+15 m (look a bit like any other numbers you've seen recently?)
f = C/ λ
f = 3.16871851e-8
T = 31558499.0467 / 3600
= 8766.249725 / 24
= 365 days

#### Ibix

You are calculating the distance you need to raise a mass $m$ in a uniform gravitational field in order for the change in gravitational potential energy to be equal to the rest mass energy of the mass.

In Newtonian physics this calculation is perfectly valid, but meaningless since "rest mass energy" isn't a concept in Newtonian physics.

In relativistic physics there are circumstances where the calculation is approximately meaningful (although your derivation via gravitational force isn't really), but it's not particularly interesting nor a general case. It's certainly nothing to do with a "space lever", whatever that may be. You are just calculating a weak-field approximation to the height difference needed for the gravitational potential change to hit a certain threshold in a particular gravitational field.

#### JSHD2019

I think in terms of the Hamiltonian (away from the classic Legrangian 1/2m.v2) actually as first posted the γm0c2 is very much more appropriate so thank you. Thence the conserved quantity H= v.p - L = γm0c2 = E; so E = γm0c2 becomes the kinetic term in the Lagrangian (Precision Cosmology). Am I still mixing apples and pears as the Americans say

#### Orodruin

Staff Emeritus
Homework Helper
Gold Member
2018 Award
Can you do the above or is the reasoning non valid and illogical? Your views
It is complete and utter nonsense.

#### Dale

Mentor
Summary: Can you cancel the masses out by turning E = mc2 into a space lever?
What is a space lever? Please provide a professional scientific reference, like a peer reviewed paper or a textbook that describes the term.

Distance x F = m.C2
Work is force times distance, not energy. Energy and work are different. They have the same units but they are not the same thing. You cannot simply substitute two quantities because they have the same units.

Distance x Acceleration = C2
The original formula, $E=mc^2$, applies for a mass at rest. So which distance and what acceleration are you talking about?

For a rest mass likely only Force acting is due to Gravity
Gravity is not a force in relativity.

Can you do the above or is the reasoning non valid and illogical? Your views
Very non valid and extremely illogical.

You seem to just be throwing around units for no purpose and seeing if something sticks. That is not logical. For each formula, you need to identify the variables with something specific in a given scenario that you are analyzing. It can be a fairly general scenario, but each variable must have an identifiable relation to a physical quantity.

There is nothing to be gained by further discussion of this incoherent assemblage of random quantities. This thread is closed.

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